Section EE  Eigenvalues and Eigenvectors

From A First Course in Linear Algebra
Version 2.00
© 2004.
Licensed under the GNU Free Documentation License.
http://linear.ups.edu/

We start with the principal definition for this chapter.

Subsection EEM: Eigenvalues and Eigenvectors of a Matrix

Definition EEM
Eigenvalues and Eigenvectors of a Matrix
Suppose that A is a square matrix of size n, x0 is a vector in n, and λ is a scalar in . Then we say x is an eigenvector of A with eigenvalue λ if

Ax = λx

Before going any further, perhaps we should convince you that such things ever happen at all. Understand the next example, but do not concern yourself with where the pieces come from. We will have methods soon enough to be able to discover these eigenvectors ourselves.

Example SEE
Some eigenvalues and eigenvectors
Consider the matrix

A = 204 98 2610 280134 36 14 716 348 9036 472232 60 28

and the vectors

x = 1 1 2 5 y = 3 4 10 4 z = 3 7 0 8 w = 1 1 4 0

Then

Ax = 204 98 2610 280134 36 14 716 348 9036 472232 60 28 1 1 2 5 = 4 4 8 20 = 4 1 1 2 5 = 4x

so x is an eigenvector of A with eigenvalue λ = 4. Also,

Ay = 204 98 2610 280134 36 14 716 348 9036 472232 60 28 3 4 10 4 = 0 0 0 0 = 0 3 4 10 4 = 0y

so y is an eigenvector of A with eigenvalue λ = 0. Also,

Az = 204 98 2610 280134 36 14 716 348 9036 472232 60 28 3 7 0 8 = 6 14 0 16 = 2 3 7 0 8 = 2z

so z is an eigenvector of A with eigenvalue λ = 2. Also,

Aw = 204 98 2610 280134 36 14 716 348 9036 472232 60 28 1 1 4 0 = 2 2 8 0 = 2 1 1 4 0 = 2w

so w is an eigenvector of A with eigenvalue λ = 2.

So we have demonstrated four eigenvectors of A. Are there more? Yes, any nonzero scalar multiple of an eigenvector is again an eigenvector. In this example, set u = 30x. Then

Au = A(30x) = 30Ax  Theorem MMSMM = 30(4x)  x an eigenvector of A = 4(30x)  Property SMAM = 4u

so that u is also an eigenvector of A for the same eigenvalue, λ = 4.

The vectors z and w are both eigenvectors of A for the same eigenvalue λ = 2, yet this is not as simple as the two vectors just being scalar multiples of each other (they aren’t). Look what happens when we add them together, to form v = z + w, and multiply by A,

Av = A(z + w) = Az + Aw  Theorem MMDAA = 2z + 2w  zw eigenvectors of A = 2(z + w)  Property DVAC = 2v

so that v is also an eigenvector of A for the eigenvalue λ = 2. So it would appear that the set of eigenvectors that are associated with a fixed eigenvalue is closed under the vector space operations of n. Hmmm.

The vector y is an eigenvector of A for the eigenvalue λ = 0, so we can use Theorem ZSSM to write Ay = 0y = 0. But this also means that y NA. There would appear to be a connection here also.

Example SEE hints at a number of intriguing properties, and there are many more. We will explore the general properties of eigenvalues and eigenvectors in Section PEE, but in this section we will concern ourselves with the question of actually computing eigenvalues and eigenvectors. First we need a bit of background material on polynomials and matrices.

Subsection PM: Polynomials and Matrices

A polynomial is a combination of powers, multiplication by scalar coefficients, and addition (with subtraction just being the inverse of addition). We never have occasion to divide when computing the value of a polynomial. So it is with matrices. We can add and subtract matrices, we can multiply matrices by scalars, and we can form powers of square matrices by repeated applications of matrix multiplication. We do not normally divide matrices (though sometimes we can multiply by an inverse). If a matrix is square, all the operations constituting a polynomial will preserve the size of the matrix. So it is natural to consider evaluating a polynomial with a matrix, effectively replacing the variable of the polynomial by a matrix. We’ll demonstrate with an example,

Example PM
Polynomial of a matrix
Let

p(x) = 14 + 19x 3x2 7x3 + x4 D = 13 2 1 02 3 1 1

and we will compute p(D). First, the necessary powers of D. Notice that D0 is defined to be the multiplicative identity, I3, as will be the case in general.

D0 = I 3 = 100 0 10 0 01 D1 = D = 13 2 1 02 3 1 1 D2 = DD1 = 13 2 1 02 3 1 1 13 2 1 02 3 1 1 = 216 5 1 0 1 87 D3 = DD2 = 13 2 1 02 3 1 1 216 5 1 0 1 87 = 19128 4 15 8 12 4 11 D4 = DD3 = 13 2 1 02 3 1 1 19128 4 15 8 12 4 11 = 749 54 5 430 4947 43

Then

p(D) = 14 + 19D 3D2 7D3 + D4 = 14 100 0 10 0 01 + 19 13 2 1 02 3 1 1 3 216 5 1 0 1 87 7 19128 4 15 8 12 4 11 + 749 54 5 430 4947 43 = 139193 166 27 98124 193 118 20

Notice that p(x) factors as

p(x) = 14 + 19x 3x2 7x3 + x4 = (x 2)(x 7)(x + 1)2

Because D commutes with itself (DD = DD), we can use distributivity of matrix multiplication across matrix addition (Theorem MMDAA) without being careful with any of the matrix products, and just as easily evaluate p(D) using the factored form of p(x),

p(D) = 14 + 19D 3D2 7D3 + D4 = (D 2I 3)(D 7I3)(D + I3)2 = 3 3 2 1 22 3 1 1 8 3 2 1 72 3 1 6 0 3 2 1 12 31 2 2 = 139193 166 27 98124 193 118 20

This example is not meant to be too profound. It is meant to show you that it is natural to evaluate a polynomial with a matrix, and that the factored form of the polynomial is as good as (or maybe better than) the expanded form. And do not forget that constant terms in polynomials are really multiples of the identity matrix when we are evaluating the polynomial with a matrix.

Subsection EEE: Existence of Eigenvalues and Eigenvectors

Before we embark on computing eigenvalues and eigenvectors, we will prove that every matrix has at least one eigenvalue (and an eigenvector to go with it). Later, in Theorem MNEM, we will determine the maximum number of eigenvalues a matrix may have.

The determinant (Definition D) will be a powerful tool in Subsection EE.CEE when it comes time to compute eigenvalues. However, it is possible, with some more advanced machinery, to compute eigenvalues without ever making use of the determinant. Sheldon Axler does just that in his book, Linear Algebra Done Right. Here and now, we give Axler’s “determinant-free” proof that every matrix has an eigenvalue. The result is not too startling, but the proof is most enjoyable.

Theorem EMHE
Every Matrix Has an Eigenvalue
Suppose A is a square matrix. Then A has at least one eigenvalue.

Proof   Suppose that A has size n, and choose x as any nonzero vector from n. (Notice how much latitude we have in our choice of x. Only the zero vector is off-limits.) Consider the set

S = x,Ax,A2x,A3x,,Anx

This is a set of n + 1 vectors from n, so by Theorem MVSLD, S is linearly dependent. Let a0,a1,a2,,an be a collection of n + 1 scalars from , not all zero, that provide a relation of linear dependence on S. In other words,

a0x + a1Ax + a2A2x + a 3A3x + + a nAnx = 0

Some of the ai are nonzero. Suppose that just a00, and a1 = a2 = a3 = = an = 0. Then a0x = 0 and by Theorem SMEZV, either a0 = 0 or x = 0, which are both contradictions. So ai0 for some i 1. Let m be the largest integer such that am0. From this discussion we know that m 1. We can also assume that am = 1, for if not, replace each ai by aiam to obtain scalars that serve equally well in providing a relation of linear dependence on S.

Define the polynomial

p(x) = a0 + a1x + a2x2 + a 3x3 + + a mxm

Because we have consistently used as our set of scalars (rather than ), we know that we can factor p(x) into linear factors of the form (x bi), where bi . So there are scalars, b1,b2,b3,,bm, from so that,

p(x) = (x bm)(x bm1)(x b3)(x b2)(x b1)

Put it all together and

0 = a0x + a1Ax + a2A2x + a 3A3x + + a nAnx = a0x + a1Ax + a2A2x + a 3A3x + + a mAmx  a i = 0 for i > m = a0In + a1A + a2A2 + a 3A3 + + a mAm x  Theorem MMDAA = p(A)x  Definition of p(x) = (A bmIn)(A bm1In)(A b3In)(A b2In)(A b1In)x

Let k be the smallest integer such that

(A bkIn)(A bk1In)(A b3In)(A b2In)(A b1In)x = 0.

From the preceding equation, we know that k m. Define the vector z by

z = (A bk1In)(A b3In)(A b2In)(A b1In)x

Notice that by the definition of k, the vector z must be nonzero. In the case where k = 1, we understand that z is defined by z = x, and z is still nonzero. Now

(A bkIn)z = (A bkIn)(A bk1In)(A b3In)(A b2In)(A b1In)x = 0

which allows us to write

Az = (A + O)z  Property ZM = (A bkIn + bkIn)z  Property AIM = (A bkIn)z + bkInz  Theorem MMDAA = 0 + bkInz  Defining property of z = bkInz  Property ZM = bkz  Theorem MMIM

Since z0, this equation says that z is an eigenvector of A for the eigenvalue λ = bk (Definition EEM), so we have shown that any square matrix A does have at least one eigenvalue.

The proof of Theorem EMHE is constructive (it contains an unambiguous procedure that leads to an eigenvalue), but it is not meant to be practical. We will illustrate the theorem with an example, the purpose being to provide a companion for studying the proof and not to suggest this is the best procedure for computing an eigenvalue.

Example CAEHW
Computing an eigenvalue the hard way
This example illustrates the proof of Theorem EMHE, so will employ the same notation as the proof — look there for full explanations. It is not meant to be an example of a reasonable computational approach to finding eigenvalues and eigenvectors. OK, warnings in place, here we go.

Let

A = 71 11 0 4 4 1 0 2 0 101 14 0 4 8 2 151 5 101 16 0 6

and choose

x = 3 0 3 5 4

It is important to notice that the choice of x could be anything, so long as it is not the zero vector. We have not chosen x totally at random, but so as to make our illustration of the theorem as general as possible. You could replicate this example with your own choice and the computations are guaranteed to be reasonable, provided you have a computational tool that will factor a fifth degree polynomial for you.

The set

S = x,Ax,A2x,A3x,A4x,A5x = 3 0 3 5 4 , 4 2 4 4 6 , 6 6 6 2 10 , 10 14 10 2 18 , 18 30 18 10 34 , 34 62 34 26 66

is guaranteed to be linearly dependent, as it has six vectors from 5 (Theorem MVSLD). We will search for a non-trivial relation of linear dependence by solving a homogeneous system of equations whose coefficient matrix has the vectors of S as columns through row operations,

3 4 6 10 18 34 0 2 6 14 30 62 3 4 6 10 18 34 5 4 2 2 10 26 4 61018 34 66  RREF 10261430 0 1371531 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

There are four free variables for describing solutions to this homogeneous system, so we have our pick of solutions. The most expedient choice would be to set x3 = 1 and x4 = x5 = x6 = 0. However, we will again opt to maximize the generality of our illustration of Theorem EMHE and choose x3 = 8, x4 = 3, x5 = 1 and x6 = 0. The leads to a solution with x1 = 16 and x2 = 12.

This relation of linear dependence then says that

0 = 16x + 12Ax 8A2x 3A3x + A4x + 0A5x 0 = 16 + 12A 8A2 3A3 + A4 x

So we define p(x) = 16 + 12x 8x2 3x3 + x4, and as advertised in the proof of Theorem EMHE, we have a polynomial of degree m = 4 > 1 such that p(A)x = 0. Now we need to factor p(x) over . If you made your own choice of x at the start, this is where you might have a fifth degree polynomial, and where you might need to use a computational tool to find roots and factors. We have

p(x) = 16 + 12x 8x2 3x3 + x4 = (x 4)(x + 2)(x 2)(x + 1)

So we know that

0 = p(A)x = (A 4I5)(A + 2I5)(A 2I5)(A + 1I5)x

We apply one factor at a time, until we get the zero vector, so as to determine the value of k described in the proof of Theorem EMHE,

(A + 1I5)x = 61 11 04 4 2 0 2 0 101 15 04 8 2 150 5 101 16 05 3 0 3 5 4 = 1 2 1 1 2 (A 2I5)(A + 1I5)x = 91 11 0 4 4 1 0 2 0 101 12 0 4 8 2 153 5 101 16 0 8 1 2 1 1 2 = 4 8 4 4 8 (A + 2I5)(A 2I5)(A + 1I5)x = 51 11 04 4 3 0 2 0 101 16 04 8 2 151 5 101 16 04 4 8 4 4 8 = 0 0 0 0 0

So k = 3 and

z = (A2I5)(A+1I5)x = 4 8 4 4 8

is an eigenvector of A for the eigenvalue λ = 2, as you can check by doing the computation Az. If you work through this example with your own choice of the vector x (strongly recommended) then the eigenvalue you will find may be different, but will be in the set 3,0,1, 1, 2. See Exercise EE.M60 for a suggested starting vector.

Subsection CEE: Computing Eigenvalues and Eigenvectors

Fortunately, we need not rely on the procedure of Theorem EMHE each time we need an eigenvalue. It is the determinant, and specifically Theorem SMZD, that provides the main tool for computing eigenvalues. Here is an informal sequence of equivalences that is the key to determining the eigenvalues and eigenvectors of a matrix,

Ax = λxAx λInx = 0 A λIn x = 0

So, for an eigenvalue λ and associated eigenvector x0, the vector x will be a nonzero element of the null space of A λIn, while the matrix A λIn will be singular and therefore have zero determinant. These ideas are made precise in Theorem EMRCP and Theorem EMNS, but for now this brief discussion should suffice as motivation for the following definition and example.

Definition CP
Characteristic Polynomial
Suppose that A is a square matrix of size n. Then the characteristic polynomial of A is the polynomial pA x defined by

pA x = det A xIn

Example CPMS3
Characteristic polynomial of a matrix, size 3
Consider

F = 1384 12 7 4 24 16 7

Then

pF x = det F xI3 = 13 x 8 4 12 7 x 4 24 16 7 x  Definition CP = (13 x) 7 x 4 16 7 x + (8)(1) 12 4 24 7 x  Definition DM + (4) 127 x 24 16 = (13 x)((7 x)(7 x) 4(16))  Theorem DMST + (8)(1)(12(7 x) 4(24)) + (4)(12(16) (7 x)(24)) = 3 + 5x + x2 x3 = (x 3)(x + 1)2

The characteristic polynomial is our main computational tool for finding eigenvalues, and will sometimes be used to aid us in determining the properties of eigenvalues.

Theorem EMRCP
Eigenvalues of a Matrix are Roots of Characteristic Polynomials
Suppose A is a square matrix. Then λ is an eigenvalue of A if and only if pA λ = 0.

Proof   Suppose A has size n.

 λ is an eigenvalue of A   there exists x0 so that Ax = λx  Definition EEM   there exists x0 so that Ax λx = 0   there exists x0 so that Ax λInx = 0  Theorem MMIM   there exists x0 so that (A λIn)x = 0  Theorem MMDAA A λIn  is singular  Definition NM det A λIn = 0  Theorem SMZD pA λ = 0  Definition CP

Example EMS3
Eigenvalues of a matrix, size 3
In Example CPMS3 we found the characteristic polynomial of

F = 1384 12 7 4 24 16 7

to be pF x = (x 3)(x + 1)2. Factored, we can find all of its roots easily, they are x = 3 and x = 1. By Theorem EMRCP, λ = 3 and λ = 1 are both eigenvalues of F, and these are the only eigenvalues of F. We’ve found them all.

Let us now turn our attention to the computation of eigenvectors.

Definition EM
Eigenspace of a Matrix
Suppose that A is a square matrix and λ is an eigenvalue of A. Then the eigenspace of A for λ, A λ, is the set of all the eigenvectors of A for λ, together with the inclusion of the zero vector.

Example SEE hinted that the set of eigenvectors for a single eigenvalue might have some closure properties, and with the addition of the non-eigenvector, 0, we indeed get a whole subspace.

Theorem EMS
Eigenspace for a Matrix is a Subspace
Suppose A is a square matrix of size n and λ is an eigenvalue of A. Then the eigenspace A λ is a subspace of the vector space n.

Proof   We will check the three conditions of Theorem TSS. First, Definition EM explicitly includes the zero vector in A λ, so the set is non-empty.

Suppose that x,y A λ, that is, x and y are two eigenvectors of A for λ. Then

A x + y = Ax + Ay  Theorem MMDAA = λx + λy  x,y eigenvectors of A = λ x + y  Property DVAC

So either x + y = 0, or x + y is an eigenvector of A for λ (Definition EEM). So, in either event, x + y A λ, and we have additive closure.

Suppose that α , and that x A λ, that is, x is an eigenvector of A for λ. Then

A αx = α Ax  Theorem MMSMM = αλx  x an eigenvector of A = λ αx  Property SMAC

So either αx = 0, or αx is an eigenvector of A for λ (Definition EEM). So, in either event, αx A λ, and we have scalar closure.

With the three conditions of Theorem TSS met, we know A λ is a subspace.

Theorem EMS tells us that an eigenspace is a subspace (and hence a vector space in its own right). Our next theorem tells us how to quickly construct this subspace.

Theorem EMNS
Eigenspace of a Matrix is a Null Space
Suppose A is a square matrix of size n and λ is an eigenvalue of A. Then

A λ = NA λIn

Proof   The conclusion of this theorem is an equality of sets, so normally we would follow the advice of Definition SE. However, in this case we can construct a sequence of equivalences which will together provide the two subset inclusions we need. First, notice that 0 A λ by Definition EM and 0 NA λIn by Theorem HSC. Now consider any nonzero vector x n,

x A λ Ax = λx  Definition EM Ax λx = 0 Ax λInx = 0  Theorem MMIM A λIn x = 0  Theorem MMDAA x NA λIn  Definition NSM

You might notice the close parallels (and differences) between the proofs of Theorem EMRCP and Theorem EMNS. Since Theorem EMNS describes the set of all the eigenvectors of A as a null space we can use techniques such as Theorem BNS to provide concise descriptions of eigenspaces. Theorem EMNS also provides a trivial proof for Theorem EMS.

Example ESMS3
Eigenspaces of a matrix, size 3
Example CPMS3 and Example EMS3 describe the characteristic polynomial and eigenvalues of the 3 × 3 matrix

F = 1384 12 7 4 24 16 7

We will now take the each eigenvalue in turn and compute its eigenspace. To do this, we row-reduce the matrix F λI3 in order to determine solutions to the homogeneous system SF λI3,0 and then express the eigenspace as the null space of F λI3 (Theorem EMNS). Theorem BNS then tells us how to write the null space as the span of a basis.

λ = 3 F 3I3 = 1684 12 4 4 24 16 4  RREF 10 1 2 0 11 2 0 0 0 F 3 = NF 3I3 = 1 2 1 2 1 = 1 1 2 λ = 1F + 1I3 = 1284 12 8 4 24 16 8  RREF 12 31 3 0 00 0 00 F 1 = NF + 1I3 = 2 3 1 0 , 1 3 0 1 = 2 3 0 , 1 0 3

Eigenspaces in hand, we can easily compute eigenvectors by forming nontrivial linear combinations of the basis vectors describing each eigenspace. In particular, notice that we can “pretty up” our basis vectors by using scalar multiples to clear out fractions. More powerful scientific calculators, and most every mathematical software package, will compute eigenvalues of a matrix along with basis vectors of the eigenspaces. Be sure to understand how your device outputs complex numbers, since they are likely to occur. Also, the basis vectors will not necessarily look like the results of an application of Theorem BNS. Duplicating the results of the next section (Subsection EE.ECEE) with your device would be very good practice.See: Computation E.SAGE.

Subsection ECEE: Examples of Computing Eigenvalues and Eigenvectors

No theorems in this section, just a selection of examples meant to illustrate the range of possibilities for the eigenvalues and eigenvectors of a matrix. These examples can all be done by hand, though the computation of the characteristic polynomial would be very time-consuming and error-prone. It can also be difficult to factor an arbitrary polynomial, though if we were to suggest that most of our eigenvalues are going to be integers, then it can be easier to hunt for roots. These examples are meant to look similar to a concatenation of Example CPMS3, Example EMS3 and Example ESMS3. First, we will sneak in a pair of definitions so we can illustrate them throughout this sequence of examples.

Definition AME
Algebraic Multiplicity of an Eigenvalue
Suppose that A is a square matrix and λ is an eigenvalue of A. Then the algebraic multiplicity of λ, αA λ, is the highest power of (x λ) that divides the characteristic polynomial, pA x.

(This definition contains Notation AME.)

Since an eigenvalue λ is a root of the characteristic polynomial, there is always a factor of (x λ), and the algebraic multiplicity is just the power of this factor in a factorization of pA x. So in particular, αA λ 1. Compare the definition of algebraic multiplicity with the next definition.

Definition GME
Geometric Multiplicity of an Eigenvalue
Suppose that A is a square matrix and λ is an eigenvalue of A. Then the geometric multiplicity of λ, γA λ, is the dimension of the eigenspace A λ.

(This definition contains Notation GME.)

Since every eigenvalue must have at least one eigenvector, the associated eigenspace cannot be trivial, and so γA λ 1.

Example EMMS4
Eigenvalue multiplicities, matrix of size 4
Consider the matrix

B = 2 1 24 12 1 4 9 6 5 24 3 4 5 10

then

pB x = 8 20x + 18x2 7x3 + x4 = (x 1)(x 2)3

So the eigenvalues are λ = 1,2 with algebraic multiplicities αB 1 = 1 and αB 2 = 3.

Computing eigenvectors,

λ = 1 B 1I4 = 3 1 24 12 0 4 9 6 5 34 3 4 5 9  RREF 10 1 3 0 0 110 0 0 0 1 0 0 0 0 B 1 = NB 1I4 = 1 3 1 1 0 = 1 3 3 0 λ = 2 B 2I4 = 4 1 24 12 1 4 9 6 5 44 3 4 5 8  RREF 10012 0 101 0 0112 0 00 0 B 2 = NB 2I4 = 1 2 1 1 2 1 = 1 2 1 2

So each eigenspace has dimension 1 and so γB 1 = 1 and γB 2 = 1. This example is of interest because of the discrepancy between the two multiplicities for λ = 2. In many of our examples the algebraic and geometric multiplicities will be equal for all of the eigenvalues (as it was for λ = 1 in this example), so keep this example in mind. We will have some explanations for this phenomenon later (see Example NDMS4).

Example ESMS4
Eigenvalues, symmetric matrix of size 4
Consider the matrix

C = 1011 0 111 1 110 1 101

then

pC x = 3 + 4x + 2x2 4x3 + x4 = (x 3)(x 1)2(x + 1)

So the eigenvalues are λ = 3,1, 1 with algebraic multiplicities αC 3 = 1, αC 1 = 2 and αC 1 = 1.

Computing eigenvectors,

λ = 3 C 3I4 = 2 0 1 1 0 2 1 1 1 1 2 0 1 1 0 2  RREF 1001 0 101 0 011 0 00 0 C 3 = NC 3I4 = 1 1 1 1 λ = 1 C 1I4 = 0011 0 011 1 100 1 100  RREF 1100 0 011 0 000 0 000 C 1 = NC 1I4 = 1 1 0 0 , 0 0 1 1 λ = 1 C + 1I4 = 2011 0 211 1 120 1 102  RREF 100 1 0 10 1 0 011 0 00 0 C 1 = NC + 1I4 = 1 1 1 1

So the eigenspace dimensions yield geometric multiplicities γC 3 = 1, γC 1 = 2 and γC 1 = 1, the same as for the algebraic multiplicities. This example is of interest because A is a symmetric matrix, and will be the subject of Theorem HMRE.

Example HMEM5
High multiplicity eigenvalues, matrix of size 5
Consider the matrix

E = 29 14 2 6 9 472211113 19 10 5 4 8 19103 2 8 7 4 3 1 3

then

pE x = 16 + 16x + 8x2 16x3 + 7x4 x5 = (x 2)4(x + 1)

So the eigenvalues are λ = 2, 1 with algebraic multiplicities αE 2 = 4 and αE 1 = 1.

Computing eigenvectors,

λ = 2 E 2I5 = 27 14 2 6 9 472411113 19 10 3 4 8 19103 4 8 7 4 3 1 5  RREF 100 1 0 0 103 21 2 0 01 0 1 0 00 0 0 0 00 0 0 E 2 = NE 2I5 = 1 3 2 0 1 0 , 0 1 2 1 0 1 = 2 3 0 2 0 , 0 1 2 0 2 λ = 1E + 1I5 = 30 14 2 6 9 472111113 19 10 6 4 8 19103 1 8 7 4 3 1 2  RREF 100 2 0 0 1040 0 01 1 0 0 00 0 1 0 00 0 0 E 1 = NE + 1I5 = 2 4 1 1 0

So the eigenspace dimensions yield geometric multiplicities γE 2 = 2 and γE 1 = 1. This example is of interest because λ = 2 has such a large algebraic multiplicity, which is also not equal to its geometric multiplicity.

Example CEMS6
Complex eigenvalues, matrix of size 6
Consider the matrix

F = 59 34 41 12 25 30 1 7 46361129 233119 58 35 75 54 157 81 43 21 5139 91 48 32 5 32 26 209 107 55 28 6950

then

pF x = 50 + 55x + 13x2 50x3 + 32x4 9x5 + x6 = (x 2)(x + 1)(x2 4x + 5)2 = (x 2)(x + 1)((x (2 + i))(x (2 i)))2 = (x 2)(x + 1)(x (2 + i))2(x (2 i))2

So the eigenvalues are λ = 2, 1, 2 + i,2 i with algebraic multiplicities αF 2 = 1, αF 1 = 1, αF 2 + i = 2 and αF 2 i = 2.

Computing eigenvectors,

λ = 2 F 2I6 = 61 34 41 12 25 30 1 5 46361129 233119 56 35 75 54 157 81 43 19 5139 91 48 32 5 30 26 209 107 55 28 6952  RREF 10000 1 5 0 1000 0 0 0100 3 5 0 00101 5 0 0001 4 5 0 0000 0 F 2 = NF 2I6 = 1 5 0 3 5 1 5 4 5 1 = 1 0 3 1 4 5

λ = 1 F + 1I6 = 58 34 41 12 25 30 1 8 46361129 233119 59 35 75 54 157 81 43 22 5139 91 48 32 5 33 26 209 107 55 28 6949  RREF 10000 1 2 0 10003 2 0 0100 1 2 0 0010 0 0 00011 2 0 0000 0 F 1 = NF + I6 = 1 2 3 2 1 2 0 1 2 1 = 1 3 1 0 1 2 λ = 2 + i F (2 + i)I6 = 61 i 34 41 12 25 30 1 5 i 46 36 11 29 233 11956 i 35 75 54 157 81 43 19 i 51 39 91 48 32 5 30 i 26 209 107 55 28 69 52 i  RREF 10000 1 5(7 + i) 0 10001 5(9 2i) 0 0100 1 0 0010 1 0 0001 1 0 0000 0 F 2 + i = NF (2 + i)I6 = 1 5(7 + i) 1 5(9 + 2i) 1 1 1 1 = 7 i 9 + 2i 5 5 5 5

λ = 2 i F (2 i)I6 = 61 + i 34 41 12 25 30 1 5 + i 46 36 11 29 233 11956 + i 35 75 54 157 81 43 19 + i 51 39 91 48 32 5 30 + i 26 209 107 55 28 69 52 + i  RREF 10000 1 5(7 i) 0 10001 5(9 + 2i) 0 0100 1 0 0010 1 0 0001 1 0 0000 0 F 2 i = NF (2 i)I6 = 1 5(7 + i) 1 5(9 2i) 1 1 1 1 = 7 + i 9 2i 5 5 5 5 So the eigenspace dimensions yield geometric multiplicities γF 2 = 1, γF 1 = 1, γF 2 + i = 1 and γF 2 i = 1. This example demonstrates some of the possibilities for the appearance of complex eigenvalues, even when all the entries of the matrix are real. Notice how all the numbers in the analysis of λ = 2 i are conjugates of the corresponding number in the analysis of λ = 2 + i. This is the content of the upcoming Theorem ERMCP.

Example DEMS5
Distinct eigenvalues, matrix of size 5
Consider the matrix

H = 15 18 8 6 5 5 3 1 1 3 0 4 5 4 2 4346 17 14 15 26 30 12 8 10

then

pH x = 6x + x2 + 7x3 x4 x5 = x(x 2)(x 1)(x + 1)(x + 3)

So the eigenvalues are λ = 2,1,0, 1, 3 with algebraic multiplicities αH 2 = 1, αH 1 = 1, αH 0 = 1, αH 1 = 1 and αH 3 = 1.

Computing eigenvectors,

λ = 2H 2I5 = 13 18 8 6 5 5 1 1 1 3 0 4 3 4 2 4346 17 16 15 26 30 12 8 12  RREF 10001 0 100 1 0 010 2 0 001 1 0 000 0 H 2 = NH 2I5 = 1 1 2 1 1

λ = 1H 1I5 = 14 18 8 6 5 5 2 1 1 3 0 4 4 4 2 4346 17 15 15 26 30 12 8 11  RREF 10001 2 0 100 0 0 010 1 2 0 001 1 0 000 0 H 1 = NH 1I5 = 1 2 0 1 2 1 1 = 1 0 1 2 2 λ = 0H 0I5 = 15 18 8 6 5 5 3 1 1 3 0 4 5 4 2 4346 17 14 15 26 30 12 8 10  RREF 1000 1 0 1002 0 0102 0 001 0 0 000 0 H 0 = NH 0I5 = 1 2 2 0 1

λ = 1H + 1I5 = 16 18 8 6 5 5 4 1 1 3 0 4 6 4 2 4346 17 1315 26 30 12 8 9  RREF 100012 0 100 0 0 010 0 0 001 12 0 000 0 H 1 = NH + 1I5 = 1 2 0 0 1 2 1 = 1 0 0 1 2 λ = 3H + 3I5 = 18 18 8 6 5 5 6 1 1 3 0 4 8 4 2 4346 17 1115 26 30 12 8 7  RREF 10001 0 100 1 2 0 010 1 0 001 2 0 000 0 H 3 = NH + 3I5 = 1 1 2 1 2 1 = 2 1 2 4 2

So the eigenspace dimensions yield geometric multiplicities γH 2 = 1, γH 1 = 1, γH 0 = 1, γH 1 = 1 and γH 3 = 1, identical to the algebraic multiplicities. This example is of interest for two reasons. First, λ = 0 is an eigenvalue, illustrating the upcoming Theorem SMZE. Second, all the eigenvalues are distinct, yielding algebraic and geometric multiplicities of 1 for each eigenvalue, illustrating Theorem DED.

Subsection READ: Reading Questions

Suppose A is the 2 × 2 matrix

A = 58 4 7

  1. Find the eigenvalues of A.
  2. Find the eigenspaces of A.
  3. For the polynomial p(x) = 3x2 x + 2, compute p(A).

Subsection EXC: Exercises

C19 Find the eigenvalues, eigenspaces, algebraic multiplicities and geometric multiplicities for the matrix below. It is possible to do all these computations by hand, and it would be instructive to do so.

C = 12 6 6

 
Contributed by Robert Beezer Solution [1200]

C20 Find the eigenvalues, eigenspaces, algebraic multiplicities and geometric multiplicities for the matrix below. It is possible to do all these computations by hand, and it would be instructive to do so.

B = 1230 5 13

 
Contributed by Robert Beezer Solution [1201]

C21 The matrix A below has λ = 2 as an eigenvalue. Find the geometric multiplicity of λ = 2 using your calculator only for row-reducing matrices.

A = 1815 33 15 4 8 6 6 9 9 16 9 5 6 9 4

 
Contributed by Robert Beezer Solution [1203]

C22 Without using a calculator, find the eigenvalues of the matrix B.

B = 21 1 1

 
Contributed by Robert Beezer Solution [1204]

M60 Repeat Example CAEHW by choosing x = 0 8 2 1 2 and then arrive at an eigenvalue and eigenvector of the matrix A. The hard way.  
Contributed by Robert Beezer Solution [1204]

T10 A matrix A is idempotent if A2 = A. Show that the only possible eigenvalues of an idempotent matrix are λ = 0 and λ = 1. Then give an example of a matrix that is idempotent and has both of these two values as eigenvalues.  
Contributed by Robert Beezer Solution [1207]

T20 Suppose that λ and ρ are two different eigenvalues of the square matrix A. Prove that the intersection of the eigenspaces for these two eigenvalues is trivial. That is, A λ A ρ = 0.  
Contributed by Robert Beezer Solution [1208]

Subsection SOL: Solutions

C19 Contributed by Robert Beezer Statement [1197]
First compute the characteristic polynomial,

pC x = det C xI2  Definition CP = 1 x 2 6 6 x = (1 x)(6 x) (2)(6) = x2 5x + 6 = (x 3)(x 2)

So the eigenvalues of C are the solutions to pC x = 0, namely, λ = 2 and λ = 3.

To obtain the eigenspaces, construct the appropriate singular matrices and find expressions for the null spaces of these matrices.

λ = 2 C (2)I2 = 32 6 4  RREF 12 3 0 0 C 2 = NC (2)I2 = 2 3 1 = 2 3

λ = 3 C (3)I2 = 42 6 3  RREF 11 2 0 0 C 3 = NC (3)I2 = 1 2 1 = 1 2

C20 Contributed by Robert Beezer Statement [1197]
The characteristic polynomial of B is

pB x = det B xI2  Definition CP = 12 x 30 5 13 x = (12 x)(13 x) (30)(5)  Theorem DMST = x2 x 6 = (x 3)(x + 2)

From this we find eigenvalues λ = 3, 2 with algebraic multiplicities αB 3 = 1 and αB 2 = 1.

For eigenvectors and geometric multiplicities, we study the null spaces of B λI2 (Theorem EMNS).

λ = 3 B 3I2 = 1530 5 10  RREF 12 0 0 B 3 = NB 3I2 = 2 1

λ = 2 B + 2I2 = 1030 5 15  RREF 13 0 0 B 2 = NB + 2I2 = 3 1 Each eigenspace has dimension one, so we have geometric multiplicities γB 3 = 1 and γB 2 = 1.

C21 Contributed by Robert Beezer Statement [1198]
If λ = 2 is an eigenvalue of A, the matrix A 2I4 will be singular, and its null space will be the eigenspace of A. So we form this matrix and row-reduce,

A2I4 = 1615 33 15 4 6 6 6 9 9 18 9 5 6 9 6  RREF 1030 0 111 0 000 0 000

With two free variables, we know a basis of the null space (Theorem BNS) will contain two vectors. Thus the null space of A 2I4 has dimension two, and so the eigenspace of λ = 2 has dimension two also (Theorem EMNS), γA 2 = 2.

C22 Contributed by Robert Beezer Statement [1198]
The characteristic polynomial (Definition CP) is

pB x = det B xI2 = 2 x 1 1 1 x = (2 x)(1 x) (1)(1)  Theorem DMST = x2 3x + 3 = x 3 + 3i 2 x 3 3i 2

where the factorization can be obtained by finding the roots of pB x = 0 with the quadratic equation. By Theorem EMRCP the eigenvalues of B are the complex numbers λ1 = 3+3i 2 and λ2 = 33i 2 .

M60 Contributed by Robert Beezer Statement [1199]
Form the matrix C whose columns are x,Ax,A2x,A3x,A4x,A5x and row-reduce the matrix,

0 6 32 102 320 966 8 10 24 58 168 490 2 12 50 156 482 1452 1 5471494791445 2 12 50 156 482 1452  RREF 1003930 0 10 1 0 1 0 01 3 10 30 0 00 0 0 0 0 00 0 0 0

The simplest possible relation of linear dependence on the columns of C comes from using scalars α4 = 1 and α5 = α6 = 0 for the free variables in a solution to SC,0. The remainder of this solution is α1 = 3, α2 = 1, α3 = 3. This solution gives rise to the polynomial

p(x) = 3 x 3x2 + x3 = (x 3)(x 1)(x + 1)

which then has the property that p(A)x = 0.

No matter how you choose to order the factors of p(x), the value of k (in the language of Theorem EMHE and Example CAEHW) is k = 2. For each of the three possibilities, we list the resulting eigenvector and the associated eigenvalue:

(C 3I5)(C I5)z = 8 8 8 24 8 λ = 1 (C 3I5)(C + I5)z = 20 20 20 40 20 λ = 1 (C + I5)(C I5)z = 32 16 48 48 48 λ = 3

Note that each of these eigenvectors can be simplified by an appropriate scalar multiple, but we have shown here the actual vector obtained by the product specified in the theorem.

T10 Contributed by Robert Beezer Statement [1199]
Suppopse that λ is an eigenvalue of A. Then there is an eigenvector x, such that Ax = λx. We have,

λx = Ax  x eigenvector of A = A2x  A is idempotent = A(Ax) = A(λx)  x eigenvector of A = λ(Ax)  Theorem MMSMM = λ(λx)  x eigenvector of A = λ2x  From this we get 0 = λ2x λx = (λ2 λ)x  Property DSAC

Since x is an eigenvector, it is nonzero, and Theorem SMEZV leaves us with the conclusion that λ2 λ = 0, and the solutions to this quadratic polynomial equation in λ are λ = 0 and λ = 1.

The matrix

10 0 0

is idempotent (check this!) and since it is a diagonal matrix, its eigenvalues are the diagonal entries, λ = 0 and λ = 1, so each of these possible values for an eigenvalue of an idempotent matrix actually occurs as an eigenvalue of some idempotent matrix. So we cannot state any stronger conclusion about the eigenvalues of an idempotent matrix, and we can say that this theorem is the “best possible.”

T20 Contributed by Robert Beezer Statement [1199]
This problem asks you to prove that two sets are equal, so use Definition SE.

First show that 0 A λ A ρ. Choose x 0. Then x = 0. Eigenspaces are subspaces (Theorem EMS), so both A λ and A ρ contain the zero vector, and therefore x A λ A ρ (Definition SI).

To show that A λ A ρ 0, suppose that x A λ A ρ. Then x is an eigenvector of A for both λ and ρ (Definition SI) and so

x = 1x  Property O = 1 λ ρ λ ρx λρ,λ ρ0 = 1 λ ρ λx ρx  Property DSAC = 1 λ ρ Ax Ax  x eigenvector of A for λρ = 1 λ ρ 0 = 0  Theorem ZVSM

So x = 0, and trivially, x 0.